Answer:
Distance is directly proportional to the velocity
Explanation:
In 1929, Edwin Hubble's wrote an article that talked about relationship between the distance and recession speed/velocity of galaxies which led to what is known as the Hubble Law. This law states that galaxies are moving away from the earth at velocities proportional to their distances.
Thus is written as;
v = H_o•d
Where;
v is velocity
d is distance
H_o is Hubble's constant rate of cosmic expansion.
He came to this conclusion by generating a graph known as Hubble's classic graph which was a graph of observed velocity vs distance for nearby galaxies.
In Newton's Third law of motion, the 'action' and 'reaction' forces act on different objects. That's why they don't cancel each other out and always result in zero force.
Answer:
F = 236063.6N
Explanation:
Please see attachment below.
Given Information:
Frequency of horn = f₀ = 440 Hz
Speed of sound = v = 330 m/s
Speed of bus = v₀ = 20 m/s
Answer:
Case 1. When the bus is crossing the student = 440 Hz
Case 2. When the bus is approaching the student = 414.9 Hz
Case 3. When the bus is moving away from the student = 468.4 Hz
Explanation:
There are 3 cases in this scenario:
Case 1. When the bus is crossing the student
Case 2. When the bus is approaching the student
Case 3. When the bus is moving away from the student
Let us explore each case:
Case 1. When the bus is crossing the student:
Student will hear the same frequency emitted by the horn that is 440 Hz.
f = 440 Hz
Case 2. When the bus is approaching the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330+20 )
f = 440 ( 330/ 350 )
f = 440 ( 0.943 )
f = 414.9 Hz
Case 3. When the bus is moving away from the student
f = f₀ ( v / v+v₀ )
f = 440 ( 330/ 330-20 )
f = 440 ( 330/ 310 )
f = 440 ( 1.0645 )
f = 468.4 Hz
Answer: B
Explanation:
Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.
The parameters given from the question are:
Mass M = 2kg
Height h = 20m
Let g = 9.8m/s^2
At the bottom of the incline plane, the object will experience maximum kinetic energy.
From conservative of energy, maximum K.K.E = maximum P.E
Maximum P.E = mgh
Maximum P.E = 2 × 9.8 × 20 = 392 J
But
K.E = 1/2mv^2
Substitute the values of energy and mass into the formula
392 = 1/2 × 2 × V^2
V^2 = 392
V = sqrt( 392 )
V = 19.8 m/s
V = 20 m/s approximately
